Question about Unitary Operators and symmetry

[hgandh]

We know that for every symmetry transformation, we can define a linear, unitary operator (or antiunitary, anti linear operator) that takes a physical state into another state. My question is if there exists unitary operators that act in this way that do not correspond to any symmetry? Would a parity transformation acting on a physical state that does not conserve it be an example of this?

 

[vanhees71]

There are a lot of unitary operators. The transformations described by them all leave the physically relevant quantities unchanged. Strictly speaking there's a larger type of transformations, not only unitary ones, because pure states are not represented by the Hilbert-space vectors but by rays in Hilbert space (or more conveniently and equivalently general quantum states by a statistical operator, with the pure states as the special case where the statistical operator is a projection operator, i.e., of the form $\hat{\rho}=|\psi \rangle \langle \psi|$ with a normalized vector $|\psi \rangle$), but one can show that any such transformation can be described either as a unitary or an antiunitary operator on Hilbert space. If the transformation can be continuously "deformed" from the identity operator, it's always a unitary transformation. The most important case, where you must (!) use an antiunitary description is "time reversal", i.e.

In detail, a unitary transformation of all Hilbert-space vectors via
$$|\psi \rangle \rightarrow |\psi ' \rangle = \hat{U} |\psi \rangle, \quad \hat{U}^{\dagger}=\hat{U}^{-1},$$
and of the operators (particularly those representing observables) via
$$\hat{A} \rightarrow \hat{A}'=\hat{U} \hat{A} \hat{U}^{\dagger},$$
leaves the probabilities according to Born's rule and also all expectation values of observables invariant:
$$\langle a' |\psi' \rangle=\langle \hat{U} a|\hat{U} \psi \rangle=\langle a |\hat{U}^{\dagger} \hat{U}|\psi \rangle=\langle a|\psi \rangle,$$
$$\langle \psi' |\hat{A}'|\psi' \rangle=\langle \hat{U} \psi |\hat{U} \hat{A} \hat{U}^{\dagger} \hat{U} |\psi \rangle=\langle \psi |\hat{U}^{\dagger} \hat{U} \hat{A} |\psi \rangle=\langle \psi |\hat{A}|\psi \rangle.$$
Also note that with $\hat{A}$ also $\hat{A}'$ is self-adjoint. Also note that $\hat{U}$ can also be time-dependent. Then one calls it a tranformation from one picture of time evolution to another (e.g., from the Schrödinger to the Heisenberg picture).

A symmetry operation is now something special. Under a symmetry we understand a transformation of quantities such that the dynamics of these quantities is unchanged. These have important physical implications. E.g., translation invariance in space means that the outcome of any experiment does not depend on where you do this experiment. E.g., investigating some heavy-ion scattering process at the LHC at CERN would give the same result as investigating the very same process at the Relativistic Heavy Ion Collider at the Brookhaven National Lab (provided you'd run the LHC at the lower energies available at RHIC ;-)).

Formally the demand on a symmetry to lead to the same dynamics of the quantum system under investigation, means that transforming the Hamiltonian with the corresponding unitary (or antiunitary) operator does not change, i.e., you must have
$$\hat{H}'=\hat{U} \hat{H} \hat{U}^{\dagger} = \hat{H},$$
or multiplying the equation with $\hat{U}$ from the right
$$\hat{U} \hat{H}=\hat{H} \hat{U} \; \Leftrightarrow \; [\hat{H},\hat{U}]=0,$$
i.e., a unitary operator represents a symmetry of the system, if it commutes with the Hamiltonian of this system.

Now, on the other hand the commutator of an operator that represents an observable, the commutator of its representing operator with the Hamiltonian represents the time derivative of this observable. Now you can express a unitary operator in terms of a self-adjoint operator via the operator exponential function,
$$\hat{U}=\exp(-\mathrm{i} \hat{A}), \quad \hat{A}^{\dagger}=\hat{A} \; \Rightarrow \; \hat{U}^{\dagger}=\hat{U}^{-1}.$$
Thus, if you have a symmetry operator $\hat{U}$, the corresponding self-adjoint "generator" $\hat{A}$ is conserved in the sense that the time derivative of the associated observable vanishes:
$$\mathring{\hat{A}}=\frac{1}{\mathrm{i} \hbar} [\hat{A},\hat{H}]=0.$$
That is the quantum version of one of Noether's theorems: Any symmetry implies the existence of a conserved quantity and vice versa.

Note that $\mathring{\hat{A}}$ is only the same as $\mathrm{d} \hat{A}/\mathrm{d} t$, if you use the Heisenberg picture to describe your system, but that's another story.

 

[hgandh]

Formally the demand on a symmetry to lead to the same dynamics of the quantum system under investigation, means that transforming the Hamiltonian with the corresponding unitary (or antiunitary) operator does not change, i.e., you must have
$$\hat{H}'=\hat{U} \hat{H} \hat{U}^{\dagger} = \hat{H},$$
or multiplying the equation with $\hat{U}$ from the right
$$\hat{U} \hat{H}=\hat{H} \hat{U} \; \Leftrightarrow \; [\hat{H},\hat{U}]=0,$$
i.e., a unitary operator represents a symmetry of the system, if it commutes with the Hamiltonian of this system.

What about Lorentz symmetry? The Hamiltonian is not left invariant in this case.

 

[bhobba]

What about Lorentz symmetry? The Hamiltonian is not left invariant in this case.

He is talking about standard QM which obeys the Galilean transformations.

See chapter 3 Ballentine.

Thanks
Bill

 

[vanhees71]

What about Lorentz symmetry? The Hamiltonian is not left invariant in this case.

I've only referred to the most simple case of symmetries, not involving explicit time dependence, which is the case for both Galilei and Lorentz boosts of course. For a full treatment of the Poincare group, see my QFT lecture notes (particularly Appendix B)

https://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf